For a signed measure $\mu$ on the Borel $\sigma$-algebra of $\mathbb{R}$ satisfying $\vert \mu \vert < \infty$ is it always possible to find a sequence of measures $\{\mu_n\}$, each a linear combination of Dirac measures, converging weakly to $\mu$? By weakly I mean

$$\int f(x) \mu_n(\text{d} x) \rightarrow \int f(x) \mu(\text{d} x)$$

as $n \rightarrow \infty$ for every bounded, continuous function $f$. If not, does it help to restrict the assumptions, e.g. to positive measures?

  • $\begingroup$ I think that the answer is affirmative as a consequence of Krein-Milman's theorem: try looking in the book by Simon. $\endgroup$ Apr 27 '18 at 10:05

Yes. First, since $||\mu||<\infty$ and $f$ is bounded, $$\int_{[-n,n]}f\,d\mu\to\int f\,d\mu.$$So you can assume that $\mu$ has compact support.

Now if $\mu$ is supported in $[-A,A]$ then $f$ is uniformly continuous on $[-A-1,A+1]$, hence $$\mu_n=\sum_j\mu([j/n,(j+1)/n))\delta_{j/n}$$works.

Edit: It's been pointed out that I'm implicitly swapping two limits in saying that we can assume $\mu$ has compact support. Above I say more or less that $\chi_{[-n,n]}d\mu\to d\mu$ weakly; in fact this convergence is in norm, and that saves the day:

For any measure $\nu$ define $$\nu_n=\sum_{j=-n^2}^{n^2}\nu([j/n,(j+1)/n))\delta_{j/n}.$$

If $\mu$ is a real measure then $\mu_n\to\mu$ weakly: Fix $\epsilon>0$. Define $$d\nu^A=\chi_{[-A,A]}d\nu.$$ If $A$ is large enough then $||\mu-\mu^A||<\epsilon$ and also $||\mu_n^A-\mu_n||<\epsilon$ for every $n$. Fix such an $A$. (Oops, the notation is ambiguous: By $\mu_n^A$ I mean $(\mu^A)_n$.)

As above we have $\lim_{n\to\infty}\int fd\mu_n^A=\int fd\mu^A$. So the triangle inequality shows that $$\left|\int fd\mu_n-\int fd\mu\right|<\epsilon(1+2||f||_\infty)$$if $n$ is large enough.

  • 1
    $\begingroup$ Thanks for the answer. I just realized, however, that I don't quite follow one part of your construction. Suppose the measure does not have compact support, then we can divide $\mathbb{R}$ into the sets $[-n,n]$ as you suggest and on each of these intervals the measures, call then $\{\mu_m\}$, will converge. If I have understood you correctly, we are left with the situation $$\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}\int_{[-n,n]}f\text{d}\mu_m = \int f \text{d}\mu,$$ but we would like the limits to be in the reverse order. How can we justify changing the order of the limits? $\endgroup$ Apr 29 '18 at 12:02
  • $\begingroup$ Fair point. In fact we can swap those limits because $\int_{[-n,n]}fd\mu\to\int fd\mu$ uniformly for $||f||\le1$; see edit for details. $\endgroup$ Apr 29 '18 at 13:31
  • $\begingroup$ Hi Ullrich, I have a question regarding the first part of your answer. You mentioned that $f$ is uniformly continuous on $[-A-1,A+1]$ but it seems like we have never used this condition(though it is correct). May I know why do we have to mention this? Thanks. $\endgroup$
    – Sam Wong
    Aug 29 '21 at 7:42
  • $\begingroup$ uniform continuity is used in the details I left out, showing that $\int f\,d\mu_n\to\int f\,d\mu$. $\endgroup$ Aug 29 '21 at 8:19
  • 1
    $\begingroup$ @SamWong How is this a problem? I_said_ that $f$ is uniformly continuous on $[-A-1,A+1]$, hence... $\endgroup$ Aug 29 '21 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.